Identify the favourable conditions for the formation of ionic bonds.
A.Low IP value of metal, low EA value of non – metal
B.High IP value of metal, high EA value of non – metal
C.Low IP value of metal, high EA value of non – metal
D.High IP value of metal, low EA value of non – metal
Answer
644.7k+ views
Hint: IP stands for Ionization Potential and EA stands for Electron Affinity.
The ionization potential or ionization energy, is the amount of energy required to remove an electron from a neutral atom or ground state.
Complete step by step answer:
When the first electron is removed from the valence shell, the energy used in the process is termed as the first ionization potential.
The basic equation for ionization energy is –
$ \Rightarrow X \to {X^ + } + {e^ - }$
The amount of energy used will change every time an electron is removed, since it becomes more difficult to remove electrons after one or more electrons has already been removed from the atom or molecule, making it positively charged.
Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom in gaseous phase in addition to an electron to the atom to form a negative ion. In other words, it is the neutral atom's tendency of gaining an electron.
$ \Rightarrow X + {e^ - } \to {X^ - }$
The amount of energy released in this process is termed as electron affinity.
For formation of an ionic bond, the low IP of metal and high EA of non – metal are the required conditions. Elements having low ionisation energies have more chances to form a cation, thus, showing a greater tendency to form ionic bonds. So, the lower ionization energy of metals favours the formation of an ionic bond. High electron affinity favours the formation of an anion. So, the elements having higher electron affinity favour the formation of an ionic bond.
Note:
The alkali metals and alkaline earth metals have a low value of IP, hence, they form cations easily. The halogens have a high value of EA, hence, they form anions easily.
Both IP and EA decrease down the group and increase across a period.
The ionization potential or ionization energy, is the amount of energy required to remove an electron from a neutral atom or ground state.
Complete step by step answer:
When the first electron is removed from the valence shell, the energy used in the process is termed as the first ionization potential.
The basic equation for ionization energy is –
$ \Rightarrow X \to {X^ + } + {e^ - }$
The amount of energy used will change every time an electron is removed, since it becomes more difficult to remove electrons after one or more electrons has already been removed from the atom or molecule, making it positively charged.
Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom in gaseous phase in addition to an electron to the atom to form a negative ion. In other words, it is the neutral atom's tendency of gaining an electron.
$ \Rightarrow X + {e^ - } \to {X^ - }$
The amount of energy released in this process is termed as electron affinity.
For formation of an ionic bond, the low IP of metal and high EA of non – metal are the required conditions. Elements having low ionisation energies have more chances to form a cation, thus, showing a greater tendency to form ionic bonds. So, the lower ionization energy of metals favours the formation of an ionic bond. High electron affinity favours the formation of an anion. So, the elements having higher electron affinity favour the formation of an ionic bond.
Note:
The alkali metals and alkaline earth metals have a low value of IP, hence, they form cations easily. The halogens have a high value of EA, hence, they form anions easily.
Both IP and EA decrease down the group and increase across a period.
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